What is the vertices ((x+5)/4)^2+((y-4)/(16))^2=1 ?

The vertices ((x+5)/4)^2+((y-4)/(16))^2=1 is (-5,20),(-5,-12)

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vertices ((x+5)/4)^2+((y-4)/(16))^2=1

vertices

(\frac{x + 5}{4})^{2} + (\frac{y - 4}{16})^{2} = 1

(- 5, 20), (- 5, - 12)

Solution steps

(h, k + b), (h, k - b)

Calculate ellipse properties

(\frac{x + 5}{4})^{2} + (\frac{y - 4}{16})^{2} = 1 :

Ellipse with center

(h, k) = (- 5, 4), b = 16, a = 4

(- 5, 4 + 16), (- 5, 4 - 16)

Refine

(- 5, 20), (- 5, - 12)

x^{2} + 25 y^{2} = 25

x^{2} + 9 y^{2} + 4 x = 5

\frac{x ^{2}}{64} + \frac{y ^{2}}{121} = 1

\frac{( x + 1 ) ^{2}}{4} + \frac{( y - 1 ) ^{2}}{9} = 1

\frac{x ^{2}}{6.25} + \frac{y ^{2}}{4.6875} = 1

What is the vertices ((x+5)/4)^2+((y-4)/(16))^2=1 ?
The vertices ((x+5)/4)^2+((y-4)/(16))^2=1 is (-5,20),(-5,-12)